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This question already has an answer here:

For $a>1, x_{n+1}=a^{x_{n}}, n=1,2, \ldots$ and $x_{1}=a$. If $\{x_{n}\}$ converges, find range of $a$.

$\begin{aligned} x_{n+1}-x_{n}&=f\left(x_{n}\right)-f\left(x_{n-1}\right)\\ &=f^{\prime}(\xi_n)\left(x_{n}-x_{n-1}\right)=\ln a e^{\xi_n \ln a}\left(x_{n}-x_{n-1}\right)\\ &=(\ln a)^2 e^{ \ln a(\xi_n+\xi_{n-1})}\left(x_{n-1}-x_{n-2}\right)\\ &\cdots\\ &=e^{(n-1)\ln\ln a+\ln a(\sum_{i=2}^{n}\xi_i)}(x_2-x_1) \end{aligned}$

If $\{x_n\}$ converges, $$e^{(n-1)\ln\ln a+\ln a(\sum_{i=2}^{n}\xi_i)}<1 \implies (n-1) \ln \ln a<-\ln a\left(\sum_{i=2}^{n} \xi_{i}\right) $$

But I can't solve this inequation.

The answer is $a \in (1,e^{\frac{1}{e}}]$.

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marked as duplicate by Simply Beautiful Art, Adrian Keister, Paul Frost, José Carlos Santos limits Sep 11 at 0:09

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Let's say that $y = x_\infty = a^{a^{a^{a^{...}}}}$. You want to find the interval in $a$ for which $y$ converges. Then $$y = a^{y}$$ We then have that $$a = y^{\frac{1}{y}}$$

The maximum of this is $e^{\frac{1}{e}}$. This can be seen by looking at $\frac{da}{dy}$: $$\frac{da}{dy} = y^{\frac{1}{y}}\left(\frac{1-\ln(y)}{y^2}\right)$$

Setting this equal to $0$, we have that $$1-\ln(y) = 0 \rightarrow y = e \rightarrow a = e^{\frac{1}{e}}$$

Therefore, $x_n$ converges for $a \in (1,e^{\frac{1}{e}}]$ if $a > 1$.

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  • $\begingroup$ This does not prove convergence for $a \leq e^{1/e}$. $\endgroup$ – Kabo Murphy Aug 2 at 23:45
  • $\begingroup$ @KaviRamaMurthy Why does it not? I'm showing that the inverse is only defined if $a < e^{\frac{1}{e}}$. $\endgroup$ – automaticallyGenerated Aug 3 at 0:06
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    $\begingroup$ You start with the asumption that $x_n$ converges to some $y$. There is no $y$ to start with . What you have proved is that if $x_n$ converges then we must have $a \leq e^{1/e}$. You have to prove the converse of this also. $\endgroup$ – Kabo Murphy Aug 3 at 0:14

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